Existence of an asymptote for $g(x)=\frac{f(x)f'(x)+f(1)f'(1)}{f'(x)+f'(1)}-f\left(\frac{xf'(x)+f'(1)}{f'(x)+f'(1)}\right)$

Question

Working with Slater's inequality (a companion of Jensen's inequality) I found this statement:

Let $f(x)$ be a continuous, twice differentiable function, convex or concave and non constant on $(0,\infty)$ and increasing on $(\alpha,\infty)$ with $\alpha>0$ a constant. Now define: $$g(x)=\frac{f(x)f'(x)+f(1)f'(1)}{f'(x)+f'(1)}-f\left(\frac{xf'(x)+f'(1)}{f'(x)+f'(1)}\right)$$ Claim: $g(x)$ has an asymptote as $x\to \infty$

I have not the key to approach the general case but let me try some examples:

For my first example I take the exponential (see here). The function can be decreasing or increasing on a such interval, see for example the function $f(x)=x^x$. As further particular cases we have $f(x)=\ln(x)$ or $f(x)=\arctan(x)$ and for these the asymptotes are constants. So far I have tried only elementary functions and their compositions but it would be curious that it works only for them.

My question:
Do you know a counter-example (it would be perfect because I have some doubt on it) or a proof (which I think is not easy) for this statement?

Thanks in advance!

Update: I cannot post it again new as a new question so I edit this with a new possible theorem:

Let $f(x)$ be a continuous, $n$ times differentiable, convex and non constant on $(0,\infty)$, increasing on $(1,\infty)$ and finally with unbounded derivatives function, then define: $$g(x)=\frac{f(x)f'(x)+f(1)f'(1)}{f'(x)+f'(1)}-f\left(\frac{xf'(x)+f'(1)}{f'(x)+f'(1)}\right)$$ With $g(x)$ strictly increasing on $(1,\infty)$.

Claim: $g(x)$ have an asymptote as $x\to \infty$.

Cross posted on math.Se

Answer 1

Here is one counter example: $f(x)=\sqrt x$ gives a function $g(x)=2-x^{1/4}-\dfrac{2}{1+\sqrt{x}}$ without an asymptote for $x\rightarrow\infty$.

Note: this is a counter example to the original conjecture in the OP, not to the updated conjecture.